# General Ring Theory/Ring extensions

Proposition (commutative ring extension is a module):

Let $R$ be a commutative ring and let $S$ be a commutative extension ring of $R$ . Then $S$ with its own addition and the restriction of the multiplication of $S$ to $R\times S$ is a module over $R$ .

Proof: From the axioms holding for rings, we deduce the module axioms as follows as follows: Let $\lambda ,\mu \in R$ and $a,b\in S$ . Then

1. $\lambda (a+b)=\lambda a+\lambda b$ (distributivity)
2. $(\lambda +\mu )a=\lambda a+\mu a$ (distributivity)
3. $\lambda (\mu a)=(\lambda \mu )a$ (commutativity of multiplication)
4. $1a=a$ (unit),

the ring axiom that's being used being indicated in the brackets. $\Box$ Proposition (Let $S/R$ be a ring extension. Then the function

$I\mapsto I\cap R$ defines a function from ideals of $S$ to ideals of $R$ .):

{{{2}}}

Proof: Indeed, $I\cap R\leq R$ because it is certainly closed under addition and multiplication by elements of $R$ . $\Box$ Proposition (Let $S/R$ be a ring extension, and suppose that $T\subseteq S$ is a multiplicative set. Then $T\cap R$ is a multiplicative set of $R$ .):

{{{2}}}

Proof: $T\cap R$ is closed under multiplication because both $T$ and $S$ are. $\Box$ 